Question

A ball of mass $50g$ is dropped from a height of $20m$. A boy on the ground hits the ball vertically upwards with a bat with an average force of $200N$, so that it attains a vertical height of $45m$. The time for which the ball remains in contact with the bat is (Take $g=10m/s^2$)

• A. 1/20th of a second
• B. 1/40th of a second
• C. 1/80th of a second
• D. 1/120th of a second

Answer 1

The correct option is B 1/40th of a second

velocity just before hitting the bat is-
$\begin{array}{c}{v}_1=\sqrt{2gh}\\ =\sqrt{2\times 10\times 20}\\ =20m/s\end{array}$
velocity of ball just leaving the bat-
$\begin{array}{c}{v}_2=\sqrt{2gh}\\ =\sqrt{2\times 10\times 45}\\ =30m/s\end{array}$
force applied by bat = rate of change of momentum
$\begin{array}{c}200=\frac{m\left(v_2-v_1\right)}{t}\\ t=\frac{m\left(30+20\right)}{200}\\ =\frac{50\times {10}^{-3}\times 50}{200}\\ =\frac{1}{40}\sec \end{array}$